Find $\lim_{x\to -3}\dfrac{\sqrt{4x+28}-4}{x+3}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2}$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
Explanation: Substituting $x=-3$ into $\dfrac{\sqrt{4x+28}-4}{x+3}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{\sqrt{4x+28}-4}{x+3} \\\\ &=\dfrac{\sqrt{4x+28}-4}{x+3}\cdot\dfrac{\sqrt{4x+28}+4}{\sqrt{4x+28}+4} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{(4x+28)-4^2}{(x+3)(\sqrt{4x+28}+4)} \\\\ &=\dfrac{4\cancel{(x+3)}}{\cancel{(x+3)}(\sqrt{4x+28}+4)} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{4}{\sqrt{4x+28}+4} \text{, for }x\neq -3 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-3$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\sqrt{4x+28}-4}{x+3}=\dfrac{4}{\sqrt{4x+28}+4}$ for all $x$ -values in the interval $(-3.5,-2.5)$ except for $x=-3$. Therefore, $\lim_{x\to -3}\dfrac{\sqrt{4x+28}-4}{x+3}=\lim_{x\to -3}\dfrac{4}{\sqrt{4x+28}+4}=\dfrac{1}{2}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -3}\dfrac{\sqrt{4x+28}-4}{x+3}=\dfrac{1}{2}$.